Problem: There is a light source on the plane. This source is so small that it can be represented as point. The light source is moving from point (a, s_y) to the (b, s_y) (s_y < 0) with speed equal to 1 unit per second. The trajectory of this light source is a straight segment connecting these two points. 

There is also a fence on OX axis represented as n segments (l_i, r_i) (so the actual coordinates of endpoints of each segment are (l_i, 0) and (r_i, 0)). The point (x, y) is in the shade if segment connecting (x,y) and the current position of the light source intersects or touches with any segment of the fence.

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You are given q points. For each point calculate total time of this point being in the shade, while the light source is moving from (a, s_y) to the (b, s_y).

Input

First line contains three space separated integers s_y, a and b (-10^9 ≤ s_y < 0, 1 ≤ a < b ≤ 10^9) — corresponding coordinates of the light source.

Second line contains single integer n (1 ≤ n ≤ 2 ⋅ 10^5) — number of segments in the fence.

Next n lines contain two integers per line: l_i and r_i (1 ≤ l_i < r_i ≤ 10^9, r_{i - 1} < l_i) — segments in the fence in increasing order. Segments don't intersect or touch each other.

Next line contains single integer q (1 ≤ q ≤ 2 ⋅ 10^5) — number of points to check.

Next q lines contain two integers per line: x_i and y_i (1 ≤ x_i, y_i ≤ 10^9) — points to process.

Output

Print q lines. The i-th line should contain one real number — total time of the i-th point being in the shade, while the light source is moving from (a, s_y) to the (b, s_y). The answer is considered as correct if its absolute of relative error doesn't exceed 10^{-6}.

Example

Input

-3 1 6
2
2 4
6 7
5
3 1
1 3
6 1
6 4
7 6


Output

5.000000000000000
3.000000000000000
0.000000000000000
1.500000000000000
2.000000000000000

Note

  * The 1-st point is always in the shade; 
  * the 2-nd point is in the shade while light source is moving from (3, -3) to (6, -3); 
  * the 3-rd point is in the shade while light source is at point (6, -3). 
  * the 4-th point is in the shade while light source is moving from (1, -3) to (2.5, -3) and at point (6, -3); 
  * the 5-th point is in the shade while light source is moving from (1, -3) to (2.5, -3) and from (5.5, -3) to (6, -3); 

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int n, m;
int sy, sx1, sx2;
int X[400010], sum[400010];
int Pointx[200010], Pointy[200010];
double K1, K2, bei;
int chaxun(double x, int Left, int Right) {
  if (Left + 1 == Right)
    return Left;
  else {
    int Mid = (Left + Right) / 2;
    if (X[Mid] < x) return chaxun(x, Mid, Right);
    if (X[Mid] >= x) return chaxun(x, Left, Mid);
  }
}
double erfen(double x1, double x2) {
  if (x2 < X[0]) return 0.00000000000;
  if (x1 > X[(2 * n) - 1]) return 0.00000000000;
  if (x1 <= X[0]) x1 = X[0];
  if (x2 >= X[(2 * n) - 1]) x2 = X[(2 * n) - 1];
  int q = chaxun(x1, 0, (2 * n) - 1), p = chaxun(x2, 0, (2 * n) - 1);
  int zuo, you;
  double whole;
  if (q % 2 == 0)
    zuo = q / 2;
  else
    zuo = (q + 1) / 2;
  if (p % 2 == 0)
    you = p / 2;
  else
    you = (p - 1) / 2;
  if (zuo == 0)
    whole = double(sum[you]);
  else
    whole = double(sum[you] - sum[zuo - 1]);
  if (q % 2 == 0) whole = whole - double(x1 - X[q]);
  if (p % 2 == 0) whole = whole - double(X[p + 1] - x2);
  whole *= bei;
  return whole;
}
int main() {
  scanf("%d%d%d", &sy, &sx1, &sx2);
  scanf("%d", &n);
  for (int a = 0; a < n; a++) {
    scanf("%d%d", &X[2 * a], &X[2 * a + 1]);
    if (a == 0)
      sum[a] = (X[2 * a + 1] - X[2 * a]);
    else
      sum[a] = sum[a - 1] + (X[2 * a + 1] - X[2 * a]);
  }
  scanf("%d", &m);
  for (int a = 0; a < m; a++) scanf("%d%d", &Pointx[a], &Pointy[a]);
  for (int a = 0; a < m; a++) {
    bei = double(Pointy[a] - double(sy)) / double(Pointy[a]);
    K1 = double(sx1 - Pointx[a]) / bei + double(Pointx[a]);
    K2 = double(sx2 - Pointx[a]) / bei + double(Pointx[a]);
    printf("%lf\n", erfen(K1, K2));
  }
  return 0;
}